Question: Define $f(x, y) = y(x + 1)$. Let $\vec{a} = (-1, -4)$ and $\vec{v} = \left( 2, 3 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Explanation: Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (-1, -4) + h \left( 2, 3 \right) \right) - f(-1, -4)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( -1 + 2h, -4 + 3h \right) - f(-1, -4)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{(-4 + 3h)(-1 + 2h + 1) - (-4)(-1 + 1)}{h}$ When we simplify, we can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{-8h + 6h^2}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{-8h + 6h^2}{h} &= \lim_{h \to 0} -8 + 6h \\ \\ &= -8 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -8$.